It is currently 26 May 2019, 07:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Jane must select three different items for each dinner she

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 6643
Followers: 107

Kudos [?]: 1274 [1] , given: 6003

Jane must select three different items for each dinner she [#permalink]  27 Sep 2017, 14:17
1
This post received
KUDOS
Expert's post
00:00

Question Stats:

54% (01:25) correct 45% (01:40) wrong based on 24 sessions

Jane must select three different items for each dinner she will serve. The items are to be chosen from among five different vegetarian and four different meat selections. If at least one of the selections must be vegetarian, how many different dinners could Jane create?

A 30

B 40

C 60

D 70

E 80
[Reveal] Spoiler: OA

_________________
Director
Joined: 20 Apr 2016
Posts: 864
WE: Engineering (Energy and Utilities)
Followers: 11

Kudos [?]: 645 [4] , given: 142

Re: Jane must select three different items for each dinner she [#permalink]  27 Sep 2017, 21:54
4
This post received
KUDOS
Carcass wrote:
Jane must select three different items for each dinner she will serve. The items are to be chosen from among five different vegetarian and four different meat selections. If at least one of the selections must be vegetarian, how many different dinners could Jane create?

A 30

B 40

C 60

D 70

E 80

there are many ways to solve the problem -((Atleast one veg means 1 Veg or 2 Veg or all 3 Veg dishes.))

SInce Jane always have 1 veg dish, we can take the following combination-

1veg dish AND 2 non-veg dishes = 5C1 * 4C2 = 5*6 = 30 ways

or 2veg dishes AND 1 non-veg dish = 5C2 * 4C1 = 10*4 = 40 ways

or 3 veg dishes = 5C3 = 10 ways

Total no. of ways Jane can create different dinners = 30+40+10 = 80 ways

Alternative way-

The number of ways to choose 3 items from the total of 9 items is 9C3 = 84.

The number of ways to choose only meat is 4C3 = 4.

Therefore, there are 84-4 = 80 different dinners
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html

Intern
Joined: 21 Nov 2018
Posts: 21
Followers: 0

Kudos [?]: 2 [0], given: 1

Re: Jane must select three different items for each dinner she [#permalink]  24 Nov 2018, 07:11
I used the following approach. Can someone tell what is wrong with this approach -:

1 veg should always be there = 5 ways in which this can be selected
now we are left with 4 veg and 4 non-veg dishes
so no of ways in which 2 items can be chosen from a total of 8 = 8C2= 28
hence total no of ways = 28*5 = 140
Supreme Moderator
Joined: 01 Nov 2017
Posts: 370
Followers: 5

Kudos [?]: 111 [0], given: 4

Re: Jane must select three different items for each dinner she [#permalink]  24 Nov 2018, 07:24
Expert's post
kunalkmr62 wrote:
I used the following approach. Can someone tell what is wrong with this approach -:

1 veg should always be there = 5 ways in which this can be selected
now we are left with 4 veg and 4 non-veg dishes
so no of ways in which 2 items can be chosen from a total of 8 = 8C2= 28
hence total no of ways = 28*5 = 140

you are going wrong because there are repetitions in it..
let me give you an example with a smaller set of numbers...

say 2 veg A and B AND 1 non veg C..
so atleast 1 veg should be there and we are looking for ways to choose two ..
By your method..
choose one of two say 2C1 and then 1 from remaining veg and 1 non veg so 1 out of two therefore 2C1
total 2C1*2C1=2*2=4
But is it so...NO
let us see the 4 ways ..
1) choose A, the other could be B or C.. AB and AC
2) choose B, the other could be A or C.. BA and BC
Isnt AB and BA same, so repetitions...

actual answer would be
both veg - 1
one veg and one non veg - 2C1*1=2
total 3

Hope it helps
_________________

Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Intern
Joined: 21 Nov 2018
Posts: 21
Followers: 0

Kudos [?]: 2 [0], given: 1

Re: Jane must select three different items for each dinner she [#permalink]  24 Nov 2018, 23:31
By my method , to chose 1 veg we have 2 ways and now we need to chose 2 items from the remaining 2 , which can be done in only 1 way or 2C2 ways.
Hence total no of dinner possible = 2 . Same as yours

Posted from my mobile device
Supreme Moderator
Joined: 01 Nov 2017
Posts: 370
Followers: 5

Kudos [?]: 111 [0], given: 4

Re: Jane must select three different items for each dinner she [#permalink]  25 Nov 2018, 00:18
Expert's post
kunalkmr62 wrote:
By my method , to chose 1 veg we have 2 ways and now we need to chose 2 items from the remaining 2 , which can be done in only 1 way or 2C2 ways.
Hence total no of dinner possible = 2 . Same as yours

Posted from my mobile device

No our answers are different, my answer here would be to choose two out of 3 =3C2=3..

Anyways the reason to tell you about that was the flaw in the approach and why you were getting wrong answers.
If you have got it, it is good.
_________________

Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Intern
Joined: 18 Feb 2019
Posts: 7
Followers: 0

Kudos [?]: 1 [1] , given: 0

Re: Jane must select three different items for each dinner she [#permalink]  21 Mar 2019, 07:13
1
This post received
KUDOS
I guess from a total of 8 = 8C2= 28 and hence total no of ways = 28*5 = 140. If you are looking for the top hot air fryer, then https://bestairfryer.reviews/gourmia/ is your choice and you have landed on the right page. It is always wise to compare the products and read the full verdict of a particular appliance. This can help you to make an unbiased decision.
Re: Jane must select three different items for each dinner she   [#permalink] 21 Mar 2019, 07:13
Display posts from previous: Sort by

# Jane must select three different items for each dinner she

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.